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How can the following:

$$ \sqrt{27-10\sqrt{2}} $$

Be simplified to:

$$ 5 - \sqrt{2} $$

Thanks

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7 Answers 7

up vote 20 down vote accepted

If you're faced with a question that says "Prove that $\sqrt{27-10\sqrt{2}}$ $=5 - \sqrt{2}$", then it's just a matter of squaring $5 - \sqrt{2}$ and seeing that you get $27-10\sqrt{2}$. But suppose the question your faced with is to find a square root of $27-10\sqrt{2}$ of the form $a+b\sqrt{2}$, where $a$ and $b$ are rational. Then you have $$ 27-10\sqrt{2}=\left(a+b\sqrt{2}\right)^2 = a^2 + 2ab\sqrt{2} + 2b^2 $$ so \begin{align} 27 & = a^2+2b^2 \\[8pt] -10 & = 2ab \end{align} From the second equation we get $a=-5/b$, then the first equation becomes $$ 27 = \frac{25}{b^2} + 2b^2 $$ or $$ 2(b^2)^2 -27b^2 + 25 = 0. $$ A solution is $b^2=1$, and you can go on from there to find $b$ and then $a$.

(And remember that the number will have two square roots.)

Later note: In order for all this to work, we have to rely on the fact that $\sqrt{2}$ is irrational. That enables us to conclude that the rational parts are equal and the irrational parts are equal, so we have two equations.

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4  
It might be worth to mention that $5-\sqrt 2>0$. –  Git Gud Mar 13 '13 at 23:31

No lucky guesses are needed, there is a simple denesting algorithm for $\rm\:\sqrt{a+b\sqrt{n}}$


Simple Denesting Rule $\rm\ \ \ \color{#0A0}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $

$\begin{array}{lll}\rm Recall\ \ \ w = a + b\sqrt{n}\ \ \ has\!\!\! &{\bf norm} &\!\!\!\rm=\: w\:\cdot\: w' = &\!\!\!\!\rm(a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 \\ \\ \rm and,\ furthermore,\ \ w\ \ has \!\!\!&{\bf trace} &\!\!\!\!\rm =\: w+w' = &\!\!\!\!\rm (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a\end{array}$


Here $\:27-10\sqrt{2}\:$ has norm $= 23^2.\,$ $\rm\, \color{#0A0}{Subtracting\ out}\ \sqrt{norm}\ = -23\ $ yields $\ 50-10\sqrt{2}\:$

and this has $\rm\ \sqrt{trace}\: =\: 10,\ \ hence\ \ \ \color{brown}{dividing\ it\ out}\ $ of this yields $\rm\ 5 - \sqrt{2} =\:$ sought sqrt.


Remark $\ $ The sign of the norm sqrt was chosen to make the trace sqrt rational. The same answer would arise using the opposite sign, but with slightly more work (rationalizing a denominator). $\ $ For many further examples see other posts on radical denesting.

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Nice technique and a good example of why there are really two solutions when taking the square root, contrary to what most mathematicians would like it to be (a single-valued function), and that it doesn't matter much whether you take the square root of an unknown or a "simple number". –  SasQ Dec 14 at 22:26
    
Where does these names come from? ("norm" and "trace") I see that the first one is similar to calculating the norm of a complex number, but there's no $i$ in it, so why does it still work similarly to complex conjugates? And why is the norm subtracted instead of divided out? Why is it the trace which is divided out instead? The trick is really nice, but it would be nicer if you explained why does this magic work so well. –  SasQ Dec 14 at 22:29
    
Oh, and one thing you wrote is a bit misleading: You put this in a way suggesting that one has to divide by the trace of the original square-rooted number ($w$), but when I tried this, it didn't work. One has to divide out by the trace of the partial answer obtained along the way instead, which is not the original $w$, but $w - \sqrt{norm}$. –  SasQ Dec 14 at 22:32

Note that $27=5^2+\sqrt2^2$ and $10\sqrt2=2\times5\times\sqrt2$

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Hint: Compare the squares of both expressions.

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Notice that $$\begin{align}27-10\sqrt{2} & = 25 - 2\cdot 5\sqrt{2} + 2 \\ & = 5^2 - 2\cdot5\sqrt2 + \left(\sqrt2\right)^2\\ & = \left(5 - \sqrt 2\right)^2 \end{align} $$

I don't know any way to notice this except to just get lucky and notice it; I didn't realize it could happen until sometime in high school when I was astounded to discover that $\sqrt{7+4\sqrt3} = 2+\sqrt3$.

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Set the nested radical as the difference of two square roots so that $$\sqrt{27-10\sqrt{2}}=(\sqrt{a}-\sqrt{b})$$ Then square both sides so that $$27-10\sqrt{2}=a-2\sqrt{a}\sqrt{b}+b$$ Set (1) $$a+b=27$$ and set (2) $$-2\sqrt{a}\sqrt{b}=-10\sqrt{2}$$ Square both sides of (2) to get $$4ab= 200$$ and solve for $b$ to get $$b=\frac{50}{a}$$ Replacing $b$ in (1) gives $$a+\frac{50}{a}=27$$ Multiply all terms by $a$ and convert to the quadratic equation $$a^2-27a+50=0$$ Solving the quadratic gives $a=25$ or $a=2$. Replacing $a$ and $b$ in the first difference of square roots formula above with $25$ and $2$ the solutions to the quadratic we have $$\sqrt{25}-\sqrt{2}$$ or $$5-\sqrt{2}$$

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What if the solutions to the quadratic were also irrational, so that $a$ and $b$ become nested inside their radicals again? Or is this possibility somehow excluded? –  SasQ Dec 14 at 22:22

$$\sqrt{27 - 10\sqrt{2}}$$ $$= \sqrt{25 + 2 - 2.5.\sqrt{2}}$$ $$= \sqrt{5^{2} - 2.5.\sqrt{2} +{\sqrt{2}}^2}$$ $$= \sqrt{{(5 - \sqrt{2})}^2}$$

$$= (5 - \sqrt{2})$$ $$= R.H.S$$

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You need equals signs rather than arrows. I would have written something similar as $$\sqrt{27-10\sqrt{2}} = \sqrt{27-2\sqrt{ 25 \times 2}} = \sqrt{25-2\sqrt{ 25 }\sqrt{2} +2} = \sqrt{(\sqrt{25} - \sqrt{2})^2}$$ –  Henry Mar 14 '13 at 8:51
    
@Henry: Thanks corrected. Its been ages I am solving some math problem –  Abhijit Mar 14 '13 at 9:13
    
But then you have to "guess" how to split that $27$. –  SasQ Dec 14 at 22:23

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